Download “Fixing” Removable Discontinuities

Making Functions Continuous
Consider: f  x  
sin x
x
Domain:
Let’s rewrite as a piecewise function
sin x


g x 
, x0
x
a,
x0
What should the value of a be such that g (x) is
continuous for all real numbers? Well, we only
know one calculus “trick” so we should
probably use it.
To be continuous at x  0 , what must be true?
Let’s try some !
x2  9
f x 
, x3
x 3
b,
x3
g ( x) 
x 2  5 x  3, x  2
ax  1,
x2
k
h ( x) 
, x  2
2
x
9  x2 , x   2
Slightly harder:
f ( x)  x 2  5 x  3, x   1
ax  b, 1  x  4
11 3 x, x  4
g ( x)  cx  1, x  3
cx 2 1, x  3
Slightly different:
x2  a2
h ( x) 
, xa
xa
2, x  a
Need more examples? Go to:
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/continuitydirectory/Continuity.html
Intermediate Value Theorem
This is an existence theorem and will not provide a
solution. It just tells us of the existence of a
solution.
If f is continuous on the closed interval  a, b 
and k is a number such that f (a)  k  f (b) ,
then there is at least one number c , such that
a  c  b , such that f (c)  k .
In calculus: Let f ( x)  x 3  x  1. Use the IVT
to show that there is at least one root [zero] on
1, 2
♪ We must use IVT, so a graph is not sufficient.
f (1)  1  1  1   1
f 2  8  2  1  5
We know that since f (x) is a polynomial, it is
continuous on the interval. [Actually, it is
continuous everywhere!] We have found that
f (1)   1 and f (2)  5 .
Our Calculus-based proof is:
By the IVT, there must exist some c , 1  c  2 ,
such that f (1)  f (c)  f (2) . Since f 1   1
and f 2  5 , then  1  f (c)  5 so there must
exist an f c   0 for c .
Construct a convincing argument using the IVT
to show that if g ( x)  x 3  5 x 2  8 x  9 , then
there exists some c such that g (c)  27 .
An example of an IVT question:
0
1
2
x
f (x)
1
2
k
The function f is continuous on [0, 2] and has
values given in the table above. The equation
1
f ( x)  has at least two solutions in the interval
2
[0, 2] if k 
1
(A)
0
(B)
(C)
1
2
(D)
2
(E)
3
Let f be a continuous function. Selected values of
f are given in the table below.
x
f x
1
2
2
3
3
1
5
5
8
7
What is the least number of solutions does the
equation f ( x) 
1 x  8
1
have on the closed interval
2
[Hint: Draw a picture first]
♪ Remember to always find the one-sided limits
for the x – value(s) in question [for #57 and
#59]. You may also want to check your solution
by graphing it.
Do page 80 # 57, 59, 60
I must clearly see the three things that make a
function continuous at a point.